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Find all solutions to the diophantine equation $7^x=3^y+4$ in positive integers. I couldn't have much progress.

Clearly $(x,y)=(1,1)$ is a solution. And there's no solution for $y=2$.

Assume $y \ge 3$ and $x \ge 1$.

By $\mod 9$, we get $7^x \equiv 4\mod 9 \implies x \equiv 2 \mod 3 $.

By $\mod 7$,we get $y \equiv 1 \mod 6$.

I also tried $\mod 2$ but it didn't work.

Please post hints ( not a solution). Thanks in advance!

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    $\begingroup$ General remark: since there is a solution, namely $(1,1)$, congruences alone won't get it done, though of course you can use congruences to eliminate a lot of cases. $\endgroup$
    – lulu
    Commented Nov 13, 2020 at 1:13
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    $\begingroup$ @lulu there is a method that usually wors with small numbers math.stackexchange.com/questions/1941354/… and many others $\endgroup$
    – Will Jagy
    Commented Nov 13, 2020 at 1:22
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    $\begingroup$ @SunainaPati FYI, your equation is $7^x - 3^y = 4$. For $x, y \gt 1$, the table in the Generalization section of Wikipedia's "Catalan's conjecture" article shows for a difference of $4$, there are only $3$ solutions with perfect powers less than $10^{18}$. These are $8 - 4 = 2^3 - 2^2$, $36 - 32 = 6^2 - 2^5$ and $125 - 121 = 5^3 - 11^2$. Thus, there's no solution for powers of $7$ and $3$ within that range, implying (but not proving) there are no other solutions than the $(1,1)$ one you've already found. $\endgroup$ Commented Nov 13, 2020 at 1:22
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    $\begingroup$ @SunainaPata Also note later in that article, Pillai's conjecture states "... each positive integer occurs only finitely many times as a difference of perfect powers ...". $\endgroup$ Commented Nov 13, 2020 at 1:24
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    $\begingroup$ @WillJagy Thanks for your feedback. I was just reading your answer for this specific equation when you commented. $\endgroup$ Commented Nov 13, 2020 at 1:29

2 Answers 2

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It's $3(3^a-1)=7(7^b-1)$ with $a=x-1$ and $b=y-1$.

Therefore $7\mid3^a-1$, so $a$ is a multiple of (what?).

Therefore, $3^a-1$ is a multiple of $13$.

Therefore, $7^b-1$ is a multiple of $13$.

Therefore, $b$ is a multiple of (what?).

Therefore, $7^b-1$ is a multiple of $9$.

Therefore, $3(3^a-1)$ is a multiple of $9$.

Therefore, $a$ is (what?).

Therefore, $x$ is (what?).

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  • $\begingroup$ Yes!! I got it :) Thankyou !!!! $\endgroup$ Commented Nov 13, 2020 at 6:49

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